At a restaurant, you can choose from 2 appetizers, 4 main courses, and 3 desserts. How many different meals, each consisting of one appetizer, one main course, and one dessert, can be created?

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13

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Apr 17, 2016

24

Explanation:

First you need to understand the process of selection and arrangement.

Suppose you have 4 numbers - 1, 2, 3, 4 and you have to form 3 digit numbers with all these numbers atmost once (For example 111 is not permitted). How would you do so?

So first step, we have to SELECT 3 numbers from this pile.

We can either choose 1,2,3 or 1,2,4 or 2,3,4 and so on. But how do you definitively know the selections. Here comes in the concept of combinations.

The basic guiding formula for combinations is that
If you have $n$ things and you have to select $r$ things, the number of selections is represented by ""^nC_r and

""^nC_r = (n!)/((n-r)! (r!))

where n! means product of all natural numbers from n to 1. (for eg. if n = 5 then n! = $5 \times 4 \times 3 \times 2 \times 1$ = 120

Here n = 4 and r = 3 so ""^nC_r = ""^4C_3 = 4 (which are 1,2,3 and 1,2,4 and 2,3,4 and 1,3,4)

Now, you need to ARRANGE these numbers. For this concept, permutations come into play, but I'll leave that for another time as it is not relevant to the question asked by you.

In your question, I just have to SELECT

1. 1 appetizer from 2 appetizers
AND
2. 1 main course from 4 main courses
AND
3. 1 dessert from 3 desserts

Notice how I have capitalized AND. It is because AND is a key word here.
Now what we have to do is get individual selection from the 3 cases and multiply them.

Remember one thing , AND means multiply and OR means add (For eg. if you had said to choose either one main course OR one dessert, then we would have added the selections)

For 1. , selections = ""^2C_1 = 2
For 2. , selections = ""^4C_1 = 4
For 3. , selections = ""^3C_1 = 3

Total possibilites = ${\text{^2C_1 xx ""^4C_1 xx }}^{3} {C}_{1} = 2 \times 4 \times 3 = 24$

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